Question

Nitrite ion${{\mathbf{NO}}^{\mathbf{-}}}_{\mathbf{2}}$, is a preservative for bacon and other foods, but it is potentially carcinogenic. A spectrophotometric determination of${{\mathbf{NO}}^{\mathbf{-}}}_{\mathbf{2}}$makes use of the following reactions

Here is an abbreviated procedure for the

determination:

1. To 50.0ml of unknown solution containing nitrite is added 1.00mL of sulfanilic acid solution.

2. After 10min , 2.00mL of 1 –a minonaphthalene solution and 1.00 mL of buffer are added.

3. After 15 min, the absorbance is read at 520 nm in a 5.00-cm cell.

The following solutions were analyzed:

A. 50.0 mL of food extract known to contain no nitrite (that is, a negligible amount); final absorbance =0.153.

B. 50.0mL of food extract suspected of containing nitrite; final absorbance \(=0.622\).

C. Same as B, but with $\mathbf{10}\mathbf{.0}\mathit{\mu }\mathbf{L}\mathbf{\text{of}}\mathbf{7}\mathbf{.50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{MNaNO}}_{\mathbf{2}}$added to the 50.0-mL sample; final absorbance =0.967.

(a) Calculate the molar absorptivity, of the colored product. Remember that a \(5.00\)-cm cell was used.

(b) How many micrograms of${{\mathbf{NO}}^{\mathbf{-}}}_{\mathbf{2}}$were present in 50.0mL of food extract?

Short answer

$o=49640{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}$

$m=4.69\mu g$

Step-by-step solution

Find molar absorptivity

1. First absorbance is calculated

A = 0.967 - 0.622 = 0.345

The concentration is:

$\begin{array}{r}\mathrm{c}=\frac{7.5\cdot {10}^{-3}\mathrm{M}\cdot 10\cdot {10}^{-6}\mathrm{L}}{0.054\mathrm{L}}\\ \mathrm{c}=1.39\cdot {10}^{-6}\mathrm{M}\end{array}$

The molar absorptivity is:

$\begin{array}{r}\dot{o}=\frac{\mathrm{A}}{\mathrm{c}\cdot \mathrm{b}}=\frac{0.345}{1.39\cdot {10}^{-6}\mathrm{M}\cdot 5\mathrm{cm}}\\ \dot{o}=49640{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}\end{array}$

Find mass

$\begin{array}{r}\mathrm{n}=7.5\cdot {10}^{-3}\mathrm{M}\cdot 10\cdot {10}^{-6}\mathrm{L}\\ \mathrm{n}=7.50\cdot {10}^{-8}\mathrm{mol}\end{array}$

of nitrite gives A=0.345.

In sample B, absorbance is:

A = 0.622 - 0.153 = 0.469

The moles of nitrite is:

$\begin{array}{c}\frac{{\mathrm{n}}_{\mathrm{x}}}{7.5\cdot {10}^{-8}\mathrm{mol}}=\frac{0.469}{0.345}\\ \mathrm{n}=\frac{0.469\cdot 7.5\cdot {10}^{-8}\mathrm{mol}}{0.345}\\ \mathrm{n}=1.02\cdot {10}^{-7}\mathrm{mol}\end{array}$

The mass can be calculated as follows

$\begin{array}{r}\mathrm{m}=\mathrm{n}\cdot \mathrm{M}=1.02\cdot {10}^{-7}\mathrm{mol}\cdot 46.01\mathrm{g}/\mathrm{mol}\\ \mathrm{m}=4.69\cdot {10}^{-6}\mathrm{g}\\ \mathrm{m}=4.69\mu \mathrm{g}\end{array}$